Temperature, States Of Matter, And Physics: Temperature Is Due To Kinetic Energy Of Particles

Temperature, States of Matter, and Physics 

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Temperature is due to Kinetic Energy of Particles – Translational – ½ mv2 – linear motion – Rotational – ½ I2 – Spinning motion – Vibrational – ½ kx2 – Spring motion In solids, vibrational rules – not enough to cause separation of particles In liquids, all three apply, but not enough to cause disruption of intermolecular forces In gases, linear rules – particles fly about freely

Thermometers and Thermal Equilibrium Thermometers work by quickly getting to the same temperature as its surroundings Thermal equilibrium is reached between two objects when they have no difference in temperature

Thermometry 

Scales of measurement – Celsius – based on boiling and freezing of water 100° boiling 0° Freezing – Fahrenheit – based on body temperature and freezing point of salt-water mixture – Conversion between C = 5/9 ( F – 32 ) or F = 1.8C + 32 Kelvin or Absolute – based on relation of volume to temperature in gases. By extrapolation, Ø K = -273.15°C

Temperature Temperature Is Not Heat Temperature measures average energy of one molecule Heat measures total energy of all molecules present

Thermal Expansion As object gains heat, molecules move farther apart Expansion vs. contraction Must be accounted for in these cases a. Dentistry b. Car engines vs piston c. Bridges and roadways

Thermal Expansion Rates different for different materials Bimetallic strips-- connection of two strips with different coefficients of expansion Motion caused by side with greater expansion or contraction Thermostats use this to act as a switch Pyrex glass made to limit expansion Liquids expand more than solids Gases expand/contract more than either solid or liquid

Thermal expansion When substances are heated, they expand Linear and Volume expansion take place Linear: Change in length is proportional to original length and change of temperature L = L0 T

 = coefficient of linear expansion

New length equation: L = L0( 1 + T )

Examples 11.3 & 11.4 The Verrazano-Narrows Bridge has a center span of 1300 m. Allowance has been made for thermal expansion and contraction of its materials, and the bridge is steel, so, for safety, allowing for a temperature range of 120°C, how much thermal expansion must be allowed for in the center span? Obtain the change in length of the center span from the definition of thermal expansion: L = L0 T

 L = (1300m)(12 x 10-6 °C-1)(120°C)

= 1.9 m. The total allowance for expansion must be 1.9 m. A copper hot-water pipe is 10.0 m long when cut and installed in a building on a day when the temperature is 10°C. How long is the pipe when it carries hot water at 60°C if the pipe is free to expand? Use the equation L = L0[1 + (T – T0)] L = (10.0 m)[1 + (17 • 10-6 °C-1)(60°C - 10°C)] L = (10.0 m)(1.00085) = 10.0085 m = 1000.85 cm.

Thermal Expansion Area and Volume: Same relation as length – Equations:  A = A0  T  V = V0  T

Gases expand most, liquids and solids much less – see table of  and  p. 319

Example 11.5 A 1.00-liter glass bottle is filled to the brim with water at a room temperature of 20 °C. The temperature of the bottle and the water is then raised to 95°C. Does the water spill over, or does the level go down, and by how much? Because the volume coefficient of thermal expansion of water changes with temperature, use the average value of  = 525 × 10-6 °C for the range of 20°C to 95°C. Write the change in volume  Vglass for the bottle as

V V

glass

= V 0  T

glass

= (27 × 10-6 °C-1)(1.00 × 10-3 m3)(95°C - 20°C) = 2.03 cm3

For the water the change in volume  Vwater is

V

V

water

= V0  T

= (525 × 10-6 °C-1)(1.00 × 10-3 m3)(95°C - 20°C) = 39.4 cm3 The expansion of the water is greater than the expansion of the bottle. The amount of water that will run over the edge is water

V

water

– Vglass = 39.4 cm3 – 2.03 cm3 = 37.4 cm3

Water: The Exception Water expands from 4°C both heating and cooling. Ice is much less dense than liquid water. Water freezes from top downward. Lakes even in very cold climates never freeze totally.

Water: The Exception Specific heat capacity is quite high Takes long to heat or to cool Cause moderation of climate of areas near large bodies of water

Gases and Physics Gases exert force on surfaces due to collision of their particles  This is translated into Pressure (force per unit area), P = F/A  Units of pressure are N/M2 or Pascals  Atmospheric gases exert pressure on everything around due to the weight and density of air at the surface 

Gas Laws Robert Boyle discovered by experiments that volume of a contained gas was inversely proportional to its pressure  Charles and Gay-Lussac likewise discovered that temperature was directly proportional to both volume and pressure  These led to the ideal gas law 

Ideal Gas Law 

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Combination of other laws involving moles, volume, temperature, and pressure Involves R, the universal gas constant has variable units, depending on usage. – – –

In physics, we use 8.31 J/mol x K Equation PV=nRT For changing conditions—Equation 

P1V1T2 = P2V2T1

Ideal Gas Law 

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Combination of other laws involving moles, volume, temperature, and pressure Involves R, the universal gas constant has variable units, depending on usage. – – –

In physics, we use 8.31 J/mol x K Equation PV=nRT For changing conditions—Equation 

P1V1T2 = P2V2T1

Kinetic Theory of Gases       

Same as in Chemistry, applies to ideal gases 1. Gases consist of many tiny particles 2. Volume of the particles is negligible compared to that of their container 3. Direction and speed of particles is entirely random 4. Collisions are perfectly elastic with no attraction between particles 5. Molecules obey Newton’s laws of motion. These apply generally but not perfectly to real gases

Kinetic Theory and Kinetic Energy 

By manipulation the Ideal Gas Law becomes – PV = 2/3(N KE) where N is the number of individual gas particles

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Rearranging and involving the temperature: – KE = 3/2 kT where k = 1.38 x 10-23 J/K – or T=2/3 KE/ k

Mechanical Equivalent of Heat Heat was thought to be a substance, caloric. Demonstrated by Count Rumford that heat happens due to work. Quantitatively demonstrated by James Prescott Joule with special apparatus Heat energy measured in both Joules and calories – calorie = amount of heat to raise 1g of water by 1°C. 1cal = 4.187 Joules

Example 11.6 A 1500-W heater is submerged into one kilogram of water that is well below 100°C. At what rate, in °C/s, does the temperature rise when the heater is operating at its rated power? The rate of energy input is 1500 J/s × 1 cal = 358.3 cal/s 4.187 J The rate of energy input per gram is 358.3 cal/s = 0.358 cal/g • s. 1000 g Since each gram of water receives 0.358 cal/s, the temperature of each gram of water, and therefore the entire volume of water, increases by 0.358°C/s.

Calorimetry  

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The measurement of heat exchanged. Depends on mass and ability to transfer heat or heat capacity. Q = mcT – Q = heat transferred – C = heat capacity

Examples 11.7 A Styrofoam cup of negligible heat capacity contains 150 g of water at 10°C. If 100 g of water at a temperature of 85°C is added, what is the final temperature of the mixture after it has been thoroughly mixed? The heat gained by the cooler water is Qgain = m1c  T1 = m1c(T – T1,0) = (150 g)c(T - 10°C). The heat lost by the hotter water is Qlost = m2c  T2 = m2c(T – T2,0) = (100 g)c(T - 85°C). When the heat lost plus the heat gained is set equal to zero, the resulting expression determines a unique value for the final temperature T: Qlost + Qgain =150c(T-10) + 100c(T-85) = 0 (150+100)T = (8500 + 1500) T= 40°C

Example 11.8 A metal block of 74 g heated in an oven to 90° C is placed in a calorimeter with 300 g water at 10°C. If the final temperature is 14°C, is the block Al, Fe, Ag, or Zn? Qlost = mcΔT = (0.074)c(14-90) =-5.62c Qgain = (0.300)(4187)(14-10) = 5024 J Qlost + Qgain = -5.62c +5024 = 0 5.62c = 5024 c= 894 J/kg · °C nearly the c of Al

Phase Change and Energy Phase change requires energy to break intermolecular forces Energy required is heat of transformation Transformation can be either at melting or boiling point. L = Q/m or Q = mL Heat of fusion at melting point Lf Heat of vaporization at boiling point L v

Example 11.9 A 105-g copper calorimeter contains 307 g of water at room temperature (T = 23ºC). If 52 g of ice at 0ºC is added to the colorimeter, what is the final temperature of the system? Qgain = miceLf + micecwaterΔT = (52 g)(80 cal/g) + (52 g)(1 cal/g • ºC)(T-0 ºC) = 4160 cal + 52T cal/ºC Qlost = mwcwΔT + mcccΔT = (mwcw + mccc)(T - T0) = [(307 g)(1 cal/g • ºC) + (105 g)(0.092 cal/g • ºC)](T-23ºC) = (317 cal/ ºC)(T-23 ºC) = 317T cal/ ºC – 7290 cal Add heat gain to heat lost 4160 cal + 52T cal/ ºC +317T cal/ ºC –7290 cal = 0 369T cal/ ºC =3130 cal T = 8.5 ºC

Heat Transfer Heat moves by three main methods: •Conduction : heat movement by direct contact of items – usually involved with solids •Convection : heat movement in fluids, where transfer is due to fluid movement •Radiation : heat movement by waves from electromagnetic sources

Conduction Happens between materials in direct contact Due to collisions of atoms or molecules and electrons Conduction accounts for cooler feel of metals vs. nonmetals Good conduction feels cool Insulators, poor conductors delay heat transfer Liquids and gases usually poor conductors/good insulators

“Cold” is not transferred

Conductivity 

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Through solids, heat conductivity is controlled by area, thickness, and temperature difference The rate of heat flow is calculated by Q/t = K • A • T /L L is thickness A is area K is thermal conductivity High values of K mean good heat conduction, low values mean good insulator.

Example 11.11 A Styrofoam cooler has surface area of 0.50 m 2 and an avg. thickness of 2.0 cm. How long will it take for 1.5 kg of ice to melt in cooler if outside temperature is 30 ºC? (thermal conductivity of Styrofoam used to make cooler is 0.030 W/m • ºC)

Q/T = KA(T

2

– T1)/L

= (0.030 W/m • ºC)(0.50 m2)(30 ºC - 0 ºC)/0.020 m = 22.5 W Heat of fusion of ice Lf = 3.34 • 105 J/kg

Q/t = m • L /t t = m • L / (Q/ t) f

f

= (1.5 kg)(3.34 • 105 J/kg)/22.5 W = 2.23 • 104 s convert to hours t =6.2 hours for all ice to melt

Insulation 

Home insulation limits heat flow and is measured accordingly – R-values listed in hardware stores tell effectiveness R = L/K

Radiation Heat due to energy waves passing thru space From electromagnetic waves of long wavelength Light also from objects emitting heat Absorption and Emission of Radiation Absorbers of radiation appear black Good absorbers also are good emitters--thermal equilibrium requires it

Radiation  Hot

objects lose energy by radiation

–Black objects fastest, silvered objects slowest  Radiated

power in watts calculated by

P = eAT4  = Stefan–Boltzmann constant (5.67 • 10 -8) e = emissivity constant for material A = area of surface Objects lose energy compared to their surroundings so this must be rewritten to Pnet =  eA(T4-Ts4)

Example 11.12 A patient waiting to be seen by his physician is asked to remove all his clothes in an examination room that is at 16°C. Calculate the rate of heat loss by radiation from the patient, given that his skin temperature is 34°C and his surface area is 1.6m2. Assume an emissivity of 0.80. Solution: The rate of heat loss by radiation is Pnet = eA (T4 - T4s), Where the temperature is expressed in kelvins. Inserting the numbers, we get Pnet = (5.67 x 10-8 W x m-2 x K-4)(0.80)(1.6 m2) [(307 K)4 – (289 K)4] Pnet = 140 W.

Review: New Equations!! C=K-273.15 K=C+273.15 F=9/5C +32 L=L0T A=A0T V=V0T 1/ mv 2 3 PV=nRT 2 rms = /2kBT vrms=√(3RT/M) Q=mcT Q=mLv or mLf

Hcond=Q/t = kA (T/L) Prad= AeT4